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How do you get the numbers 33-99 by using the numbers 2,0,2,4

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  • $\begingroup$ Does this answer your question? Make numbers 1 - 30 using the digits 2, 0, 2, 4 $\endgroup$
    – Someone
    Mar 1 at 3:22
  • $\begingroup$ There are already quite a few questions almost exactly like this—while the range is different, it's fundamentally the same thing being asked. $\endgroup$
    – Someone
    Mar 1 at 3:23
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    $\begingroup$ These kind of puzzles require a list of allowed operations. Exactly what are we allowed to do? Addition? Division? Exponentiation? Concatenation? Square roots? N-th roots? Floor? Ceiling? Factorial? Trigonometry? Parentheses? And so forth. $\endgroup$
    – bobble
    Mar 1 at 4:44

1 Answer 1

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I'm assuming we can use exponentiation, factorial, and N-th root operations
Here are my solutions so far -->
(Sorry for the bad formatting!)

33 = (2 + 0!)2 + 4!
34 = 2 $\times$ (0! + 42)

35 = $\dfrac{4 + 2 + 0!}{.2}$

36 = (2 + 0!)! $\times$ (2 + 4)
37 = (4 + 2)2 + 0!
38 = ?
39 = $\left(\dfrac{4}{.2} \times 2 \right) - 0!$

40 = $\left(\dfrac{4}{.2} \times 2 \right) + 0$

41 = $\left(\dfrac{4}{.2} \times 2 \right) + 0!$

42 = 2 $\times$ (4! - (2 + 0!))
43 = 2 $\times$ (4! - 2) - 0!
44 = (2 $\times$ 4!) - (0 $\times$ 2)

45 = $\dfrac{(4 \times 2) + 0!}{.2}$

46 = (4! - (0 $\times$ 2)!) $\times$ 2
47 = ?
48 = 42 $\times$ (2 + 0!)
49 = (2 + 4 + 0!)2
50 = (4 + 0!)2 $\times$ 2
51 = ?
52 = (4! + 2) $\times$ (2 + 0)
53 = ((4! + 2) $\times$ 2) + 0!
54 = ?
55 = ?
56 = ?
57 = ?
58 = ?

59 = $\dfrac{4!}{2 \times .2} - 0!$

60 = $\dfrac{2 + 4}{.2} \times 2$

61 = $\dfrac{4!}{2 \times .2} + 0!$

62 = 42 + 0! - 2
63 = 22 + 4 - 0!
64 = 22 + 4 + 0
65 = 22 + 4 + 0!
66 = (4! - 2) $\times$ (2 + 0!)
67 = ?
68 = ?
69 = ?
70 = ?
71 = ?
72 = ?
73 = ?
74 = ?

75 = $\dfrac{4^2 - 0!}{.2}$

76 = ?
77 = ?
78 = ?

79 = $\dfrac{4^2}{.2} - 0!$

80 = $\dfrac{4^2}{.2} + 0$

81 = $\dfrac{4^2}{.2} + 0!$

82 = ?
83 = (2 + 0!)4 + 2
84 = ?

85 = $\dfrac{4^2 + 0!}{.2}$

86 = ?
87 = ?
88 = ?
89 = ?
90 = ?
91 = ?
92 = (4! - 0!) $\times$ 2 $\times$ 2
93 = ?
94 = ?

95 = $\dfrac{\dfrac{4}{.2} - 0!}{.2}$

96 = ?
97 = ?
98 = ?

99 = $\dfrac{4}{.2 \times .2} - 0!$

I haven't figured out the answer for the numbers with the question marks yet.
I'll try to update as soon as I get the answers

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